91Ó°ÊÓ

Four points that are vertices of a quadrilateral are given: $(1,-1),(4,1),(2,2)$and $(5,4)$.

Show that the given quadrilateral is a parallelogram.

Slope, $m=\frac{y_2-y_1}{x_2-x_1}$

Slope of line joining $(1,-1)$and $(4,1)$is: $m_1=\frac{1-(-1)}{4-1}$

$m_1=\frac{2}{3}$

Slope of line joining localid="1646095842603" $(2,2)$and localid="1646095846511" $(5,4)$is: localid="1646095799965" $m_2=\frac{4-2}{5-2}$

$m_2=\frac{2}{3}$

Slope of line joining localid="1646095850881" $(4,1)$and localid="1646095855754" $(5,4)$is: localid="1646095805288" $m_3=\frac{4-1}{5-4}$

$m_3=3$

Slope of line joining localid="1646095861870" $(1,-1)$and localid="1646095867258" $(2,2)$is: localid="1646095812202" $m_4=\frac{2-(-1)}{2-1}$

$m_4=3$

As $m_1=m_2$and $m_3=m_4$, the slopes of opposite sides of quadrilateral are equal. So, the lines are parallel.

Therefore, given quadrilateral is a parallelogram.